Answer
$5.68\times10^{-24}\;N$
Work Step by Step
The electrostatic force on particle 4 due to the particle 1 is
$F_{41}=\frac{1}{4\pi\epsilon_0}\frac{3\times10^{-19}\times3\times10^{-19}}{0.03^2}\;N=9\times10^{-25}\;N$, directed along $\vec{d_1}$ towards the center
The positive x-component of $\vec{F_{41}}$ is
$F_{41_x}=9\times10^{-25}\cos35^{\circ}\;N=7.37\times10^{-25}\;N$
The positive y-component of $\vec{F_{41}}$ is
$F_{41_y}=9\times10^{-25}\sin35^{\circ}\;N=3.80\times10^{-25}\;N$
The electrostatic force on particle 4 due to the particle 2 is
$F_{42}=\frac{1}{4\pi\epsilon_0}\frac{3\times10^{-19}\times3\times10^{-19}}{0.02^2}\;N=2.025\times10^{-24}\;N$, directed along positive y-axis
The electrostatic force on particle 4 due to the particle 3 is
$F_{43}=\frac{1}{4\pi\epsilon_0}\frac{3\times10^{-19}\times6.40\times10^{-19}}{0.02^2}\;N=4.32\times10^{-24}\;N$, directed along positive x-axis
Therefore, total electrostatic force on particle 4 along positive x-axis is
$F_x=(7.37\times10^{-25}+4.32\times10^{-24})\;N=5.057\times10^{-24}\;N$
total electrostatic force on particle 4 along positive y-axis is
$F_y=(3.80\times10^{-25}+2.205\times10^{-24})\;N=2.585\times10^{-24}\;N$
Therefore, the net electrostatic force on particle 4 due to the other three
particles is
$F=\sqrt {F_x^2+F_y^2}$
or, $F=\sqrt {(5.057\times10^{-24})^2+(2.585\times10^{-24})^2}\;N$
or, $F=5.68\times10^{-24}\;N$
